結果
問題 | No.694 square1001 and Permutation 3 |
ユーザー |
|
提出日時 | 2019-05-12 14:58:24 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 2,782 ms / 3,000 ms |
コード長 | 4,060 bytes |
コンパイル時間 | 1,977 ms |
コンパイル使用メモリ | 183,264 KB |
実行使用メモリ | 128,256 KB |
最終ジャッジ日時 | 2024-07-04 13:13:11 |
合計ジャッジ時間 | 15,674 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 13 |
ソースコード
// includes#include <bits/stdc++.h>// macros#define ll long long int#define pb emplace_back#define mk make_pair#define pq priority_queue#define FOR(i, a, b) for(int i=(a);i<(b);++i)#define rep(i, n) FOR(i, 0, n)#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)#define vrep(v, i) for(int i = 0; i < (v).size(); i++)#define all(x) (x).begin(),(x).end()#define sz(x) ((int)(x).size())#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())#define FI first#define SE second#define dump(a, n) for(int i = 0; i < n; i++)cout << a[i] << "\n "[i + 1 != n];#define dump2(a, n, m) for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)cout << a[i][j] << "\n "[j + 1 != m];using namespace std;// typestypedef pair<int, int> P;typedef pair<ll, int> Pl;typedef pair<ll, ll> Pll;typedef pair<double, double> Pd;typedef complex<double> cd;// constantsconst int inf = 1e9;const ll linf = 1LL << 50;const double EPS = 1e-10;const int mod = 1e9 + 7;// solvetemplate <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr =itr; if(++titr != st.end())os << " ";} return os;}template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr;auto titr = itr; if(++titr != st.end())os << " ";} return os;}template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; autotitr = itr; if(++titr != st.end())os << " ";} return os;}template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os <<*itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << p.first << " " << p.second; return os;}template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os <<itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end();++itr){ os << itr->first << ":" << itr->second; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}template<typename T>struct BIT_{int n;vector<T> bit;BIT_(int n_){n = n_;bit = vector<T>(n+1, 0);}void add(int a, T w){for(int x = a; x < n; x |= (x + 1)){bit[x] += w;}}T sum(int a){T res = 0;for(int x = a; x >= 0; x = (x & (x + 1)) - 1){res += bit[x];}return res;}};typedef struct BIT_<int> BITI;typedef struct BIT_<long long> BITL;int main(int argc, char const* argv[]){ios_base::sync_with_stdio(false);cin.tie(0);int n;cin >> n;vector<ll> a(n);cin >> a;BITL bt(n);map<ll, vector<int>> mp;rep(i, n)mp[a[i]].pb(i);ll res = 0;irep(itr, mp){for(auto e: itr->second){res += bt.sum(n - 1) - bt.sum(e);}for(auto e: itr->second){bt.add(e, 1);}}int at = 0;map<ll, int> tr, tr2;irep(itr, mp){tr[itr->first] = at;at += sz(itr->second);tr2[itr->first] = at;}cout << res << endl;rep(i, n - 1){res = res - tr[a[i]];res += (n - tr2[a[i]]);cout << res << endl;}return 0;}